Calorimetry - Chemistry

Question

A 58.78 gram sample of iron (with a heat capacity of 0.450 ^J/_gºC) is heated to 100.0 ºC. It is then transferred to a coffee cup calorimeter containing 33.63 g of water (specific heat of 4.184 ^J/ _gºC) initially at 20.63 ºC. If the final temperature of the system is 23.59ºC, how much heat was absorbed by the bomb calorimeter? (Please give the absolute value) 

Answer 1

Heat lost by the hot iron equals heat gained by the water PLUS heat gained by the calorimeter.

heat lost by iron = q = mC∆T

q = (58.78 g)(0.450 J/gº)(100º - 23.59º) = (58.78 g)(0.450 J/gº)(76.41º) = 2021 J

heat gained by water = q = mC∆T

q = (33.63 g)(4.184 J/gº)(23.59º - 20.63º) = (33.63 g)(4.184 J/gº)(2.96º) = 416 J

heat gained by calorimeter = heat lost by iron - heat gained by water

heat gained by calorimeter = 2021 J - 416 J = 1605 J absorbed by the calorimeter.